On Tuesday 26 February 2002 20:50, Andrew Slezak wrote:
> Basically I am setting ls > array(). I want each item to be an option in
> the array. I have have scoured for some time now and this is what I have
> come up with, but it is not populating each file as an <option>? The
> select box is just empty. I know this is an array because I have tested
> with print_r. Stuck!
>
> Can someone tell me what I am doing wrong?
> ------------------------------------------------
> <?
> $shell_scripts=`ls -1 /sm_scripts|grep -v backup|grep -v outmail|grep -v
> c_`;
> $shell_scripts=rtrim($shell_scripts);
> $arr1=split("\n", "$shell_scripts");
> //print_r($arr1);
>
>
> echo "<p>";
> echo "<select name=test>";
> function generate_option_list($arr1, $set) {
> reset($arr1);
> while (list($key,$value) = each($arr1) ) {
> {$var.="<option value='$key'>$value</option>";}
> } return $var;
> }
>
> echo "</select>";
> ------------------------------------------------
> ?>
You've defined a function "generate_option_list", but you haven't called it!
--
Jason Wong -> Gremlins Associates -> www.gremlins.com.hk
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